|Hypothesis testing homework solutions
Setting up and interpreting results of a hypothesis test
ACT-1. Practice with p-values
Use the density tool to practice finding p-values.
I. Set the mean to 35, the standard deviation to 12.5, and the sample size to 100. What is the probability of finding a value:
a. less than 33 .055
b. greater than 37.1 .039
c. less than 32.8 or greater than 36.9 .039+.065=.104
II. Change the sample size to 250. Now that the sample size is greater, find the following probabilities and compare them with those found in the first question:
a. less than 33 .006
b. greater than 37.1 .004
c. less than 32.8 or greater than 36.9 .003+.008=.011
Larger samples taken from the same population have a much smaller chance to have a mean value far from the population mean. Increasing the sample size by a factor of 2.5 decreased the likelihood of these rare events by a factor of 10.
III. Keeping the sample size at 250 and the mean at 35, try changing the standard deviation to 11 (a difference of only 1.5 from the previous standard deviation) and compare these results with the results you obtained in the second question.
a. less than 33 .034
b. greater than 37.1 .029
c. less than 32.8 or greater than 36.9 .023+.042=.065
A relatively small, about 10% decrease in the standard deviation had a huge effect increasing (by a factor of 6) the chances for certain sample mean values to come up that are far form the population mean..
IV. Set the mean to 34, the standard deviation to 12.5, and the sample size to 100. Find the following probabilities and compare them with the ones you found in the first question.
a. less than 33 .21
b. greater than 37.1 .007
c. less than 32.8 or greater than 36.9 .167+.01=.177
33 is much closer (almost 1.25 closer which would be a full standard deviation closer) to the mean now so chances for a sample mean to be 33 increased dramatically. On the other and 37.1is a full standard deviation further than where it was when the mean was 35. therefore fewer samples show this value. In part c) the lower value got closer the higher value mover further form the mean so it evens out to about the same chance overall as in the corresponding part in I.
Write a brief summary of what you observed when certain values were changed.
MRA-1. Population Mean Hypotheses
Each of the following paragraphs calls for a statistical test about a population mean m. State the null hypothesis H0 and the alternative hypothesis Ha in each case.
(a)The diameter of a spindle in a small motor is supposed to be 5 mm. If the spindle is either too small or too large, the motor will not work properly. The manufacturer measures the diameter in a sample of motors to determine whether the mean diameter has moved away from the target.
(b)Census Bureau data show that the mean household income in the area served by a shopping mall is $42,500 per year. A market research firm questions shoppers at the mall. The researchers suspect the mean household income of mall shoppers is higher than that of the general population.
©The examinations in a large accounting class are scaled after grading so that the mean score is 50. The professor thinks that one teaching assistant is a poor teacher and suspects that his students have a lower mean score than the class as a whole. The TA's students this semester can be considered a sample from the population of all students in the course, so the professor compares their mean score with 50.
MRA-4. Explaining Ethnocentrism
A social psychologist reports that in our sample, "ethnocentrism was significantly higher (p= 0.05) among church attenders than among nonattenders."
Explain what this means in language understandable to someone who knows no statistics. Do not use the word "significance" in your answer.
Researchers found that the percentage of ethnocentrics was higher among church attenders compared with the same percentage for non-attenders. In fact the evidence for that was so extreme that if there was no difference between the percentage of ethnocentrics between attenders and non-attenders we would see such an extreme result only 5% of the time, or 5 out of 100 times.
MRA-5. Explain p-value
When asked to explain the meaning of "the p-value was p = 0.03," a student says, "This means there is only probability 0.03 that the null hypothesis is true." Is this an essentially correct explanation? Explain your answer.
This is a false explanation. We do not know if H0 is true or false. The p-value tells us that if Ho was true we only had 3% chance of seeing the evidence that we collected. That means it is very unlikely to see our evidence if H0 is true but it does not tell us what is the chance for Ho to be true.
MRA-6. Explain Significance Again
When asked why statistical significance appears so often in research reports, a student says, "Because saying that results are significant tells us that they cannot easily be explained by chance variation alone." Do you think that this statement is essentially correct? Explain your answer.
This is essentially correct. A result is significant evidence against H0 if there was a very small chance to see it if H0 was true. So a result is significant if you cannot easily say that it could have happened simply by chance that you got this extreme result even though Ho was true.
MRB-4. Students' earnings
The financial aid office of a university asks a sample of students about their employment and earnings. The report says that "for academic year earnings, a significant difference (P = 0.038) was found between the sexes, with men earning more on the average. No difference (P = 0.476) was found between the earnings of black and white students." Explain both of these conclusions, for the effects of sex and of race on mean earnings, in language understandable to someone who knows no statistics.
The evidence for men earning more than women was so strong that by chance you would see this evidence only 3.8% of the times. So we will conclude that the reason is not chance but that there truly is a difference between the earnings of men and women.
In the second case there was no compelling evidence since we could have seen this evidence even if there was no difference between the earnings of whites and blacks
47 out of 100 times. So we had almost 50% chance to see our evidence if there was no difference. We will conclude that we have no significant evidence to believe that earnings are different.
Testing a Hypothesis
MRA-3. Market Pioneers
Market pioneers, companies that are among the first to develop a new product or service, tend to have higher market shares than latecomers to the market. What accounts for this advantage? Here is an excerpt from the conclusions of a study of a sample of 1209 manufacturers of industrial goods:
"Can patent protection explain pioneer share advantages? Only 21% of the pioneers claim a significant benefit from either a product patent or a trade secret. Though their average share is two points higher than that of pioneers without this benefit, the increase is not statistically significant (z = 1.13). Thus, at least in mature industrial markets, product patents and trade secrets have little connection to pioneer share advantages."
Find the p-value for the given z. Then explain to someone who knows no statistics what "not statistically significant" in the study's conclusion means. Why does the author conclude that patents and trade secrets don't help, even though they contributed 2 percentage points to average market share?
To find the p-value we look up z=1.13 in the normal table. Now the chances to see our evidence or an even more compelling evidence with z>1.13 equals to the area to the right of 1.13. That is close to 13%. So assuming that there was no real difference between the market shares between companies with product patents and those without them, we still have 13% chance of seeing the 1209 companies investigated doing that much better than the others. Therefore it is not unlikely at all that these companies were doing better simply by chance of selection and not because they had a product patent. So a 2% point differenc in market share can occur within samples of 1209 quite easily and we do not have the basis for rejecting the possibility that trade secrets and product patents make no difference.
MBS-4. One-year Returns on Electric Utility Stocks
The one-year rate of return to shareholders in 1995 was calculated for each in a sample of 63 electric utility stocks. The data, extracted from The Wall Street Journal are recorded in the attached dataset.
1. Specify the null and alternative hypotheses tested for determining whether the true mean one-year rate of return for electric utility stocks exceeded 30%.
2.Calculate the observed significance level of the test.
Go to Calculate/Test choose t-test for individual means and Ho: µ=.3, Ha:µ>0.3. The p-value is reported to be less than .0001 which is indeed super small.
3.Interpret the result in #2 in the context of the problem.
We have strong evidence that makes us believe that the average one year return for electric utility stocks exceeds 30%. We will reject H0 and accept the alternative hypothesis.
MCS-4. Contaminated Soil and p-values
Environmental Science & Technology (Oct. 1993) reported on a study of contaminate soil in The Netherlands. Seventy-two(n=72) 400-gram soil specimens were sampled, dried, and analyzed for the contaminant cyanide. The cyanide concentration [in milligrams per kilogram (mg/kg) of soil] of each soil specimen was determined using an infrared microscopic method. The sample resulted in a mean cyanide level of x-bar = 84mg/kg and a standard deviation of s = 80 mg/kg.
Test the hypothesis that the true mean cyanide level in soil in The Netherlands exceeds 100 mg/kg. Use a = .10.
Ho: µ=100 mg/kg
Ha: µ>100 mg/kg
You have to calculate what is the probability of a sample to have a mean 84 or larger if the population mean truly was 100. So you set the mean=100, n=72 , s=80 and find the area to the right of 84. And we find that we have a 95.55% chance to see such a sample if it came from a population with mean 100. Therefore there is no significant evidence to believe that the
Mean contamination is higher than 100. If anything we would tend to believe based on this evidence that the mean contamination was lower than 100.
Would you reach the same conclusion as in part a using a = .05? Using a =.01? . At alpha= .10, .05 or.01 we would equally would not be able to reject Ho since our p-value is way above any of these alpha levels..
Why can the conclusion of a test change when the value of p=alpha is changed? It could change if the p-value would be say .03. At alpha -.1 or .05 it is considered small enough to lead us to rejection but at alpha=.01 level it is not significant because a smaller alpha level requires evidence that has less than .01 probability of occurring.
Therefore at the .01 level we could not reject H0 based on this evidence with p-value=.03.
MCS-6. Defective Products
A large mail-order company has placed an order for 5,000 electric can openers with a supplier on the condition that no more than 2% of the devices will be defective. To check the shipment, the company tests a random sample of 400(n=400)of the can openers and finds that 11 (sample proportion=11/400=.0275)are defective.
1.Does this provide sufficient evidence to indicate that the proportion of defective can openers in the shipment exceeds 2%? Test using a = .05.
2. sample proportion is .0275 and n=400.
3.Use density tool click on the box and choose proportion. Enter p=.02, n=400 and move the flag to find the area to the right of .0275. the area is .147 meaning that this sample had a 14.7% chance to come from this distribution where the mean is in fact 2%. At alpha =.05 level there is no ground to reject the hypothesis that the mean is in fact 2%.
Find and interpret the observed significance level for the hypothesis test.
P=.147 there is a 14.7% chance that we have seen this slightly higher value in this shipment of 400 can openers by chance.
These data are the total weights of garbage discarded by households in one week (based on data collected as part of the Garbage Project at the University of Arizona). At the 0.01 level of significance, test the claim of the city of Providence supervisor that the mean weight of all garbage discarded by households each week is less than 35 lb., the amount that can be handled by the town. Based on the result, is there any cause for concern that there might be too much garbage to handle?
Calculate/test. Choose individual t-test, µ=35 Ha: µ<35. Set alpha=.001. We got p<.0001 so we have significant evidence to reject Ho and accept the supervisors claim that in fact the average household trash is less than 35 lb. there is no cause for concern based on the evidence.
WEN-4. Radio Ownership
The Radio Advertising Bureau of New York reports in Radio Facts, that in 1994 the mean number of radios per U.S. household was 5.6. A random sample of 45 U.S. households taken this year yields the data recorded in Data Desk on number of radios owned.
Does the data provide sufficient evidence to conclude that this year's mean number of radios per U.S. households has changed from the 1994 mean of 5.6? Use the following steps to answer the question.
1.State the null and alternative hypothesis. H0: µ=5.6, Ha: µ≠5.6
2.Discuss the logic of conducting the hypothesis test.
We will use the statistic form the sample to calculate how likely it is to see a value of a sample mean differ from the population mean this much or more either being that much larger or that much smaller. The p-value thus will account for possible chance differences of this magnitude in both tails or directions.
2.Identify the distribution of the variable x, that is, the sampling distribution of the mean for samples of size 45. the sampling distribution will be normal since below the population standard deviation is given. The mean for the sampling distribution is 5.6 and it's stndsrd deviation is 1.9/square root of 45.
3. Obtain a precise criterion for deciding whether to reject the null hypothesis in favor of the alternative hypothesis.
Let us set alpha=.05. There is no particular danger associated with a type I error so alpha can be that large.
Apply the criterion in the previous question to the sample data and state your conclusion. Assume the population standard deviation of the year's number of radios per U.S. households is 1.9.
Calculate /test use z-intervals, specify sigma and mu. You get p=.30 and fail to reject H0 obviously. No evidence to believe that there is any difference in the mean number of radios per household.
WEN-6. Hospital Costs
The American Hospital Association reports in Hospital Stat that the mean cost to community hospitals per patient per day in U.S. hospitals was $931 in 1994. In that same year, a random sample of 30 daily costs in Massachusetts hospitals yielded a mean of $1131. Assuming a population standard deviation of $333 for Massachusetts hospitals, do the data provide sufficient evidence to conclude that in 1994 the mean cost in Massachusetts hospitals exceeded the national mean of $931? Perform the required hypothesis test at the 5% significance level.
2. xbar-1131, n=30, sigma=333
3. Use density tool. Choose mean. Enter Mean=931, s=333, n=30 if you move your flag it leaves the screen at 1113 anything above that value has a p-value less then .0001in particular our sample mean 1131 has p-value less than.0001 consequently. We will reject H0 at 5% level. *Finish Interpretation….
WEN-8. The Focus Database
The Focus database contains information on 500 randomly selected Arizona State University sophomores. Seven variables are considered for each student: sex, high-school GPA, SAT math score, cumulative GPA, SAT verbal score, age and total hours. Use a statistical software package solve the following problems.
1.In 1995 the (no-recentered) mean SAT math score was 482 nationally. Do the Focus data on SAT math scores provide sufficient evidence to conclude that the mean SAT math score of Arizona State University sophomores exceeds the 1995 national mean? Use alpha equal to 0.05.
yes, p<0.001 so we will reject H0 at 55 level.
2. In 1995 the (no-recentered) mean SAT verbal score was 482 nationally. Do the Focus data on SAT verbal scores provide sufficient evidence to conclude that the mean SAT verbal score of Arizona State University sophomores exceeds the 1995 national mean? Use alpha equal to 0.05.
fail to reject, p=1.oo this low score of 456 does not provide evidence obviously for the average being higher than 482.
3. Justify the use of the hypothesis-testing procedures that you employed in parts 1 and 2. Scores in the samples were normally distributed, no outliers that makes our assumption that the population is normal plausible.
In a discussion of the education level of the American Workforce, someone says "The average young person can't even balance a checkbook." The National Assessment of Educational Progress (NAEP) says that a score of 275 or higher on its quantitative test reflects the skill needed to balance a checkbook. The NAEP random sample of 840 young men had a mean score of x-bar = 272, a bit below the checkbook-balancing level. Assume that sigma = 60, choose an alpha level, and perform a hypothesis test to determine whether this is good evidence to conclude that the mean score for all young men is less than 275.
2. xbar=272, sigma=60, n=840 sigma of xbar=60/root840=2
4. Since the p-value is higher than my set alpha level I will not reject the null hypothesis. We have no significant evidence to believe that young men can't balance their checkbooks based on this sample.
Type I , Type II error
YMM-3. Testing for ESP
A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P less than .01) than random guessing.
Based on this, is it proper to conclude that these four people have ESP? Explain.
What should the researcher do now to test whether any of these four subjects have ESP?
A test at 1% level has a full 1% chance to reject H0 in favor of Ha. So if you repeated the test 500 times you expect this type I error to occur about 5 out of the 500 times. So seeing rejection 4 out of 500 times could be explained by these being the incidents for a Type I error and therefore 4 out of 500 rejections will give no evidence for ESP.
A group of psychologists once measured 77 variables on a sample of schizophrenic people and a sample of people who were not schizophrenic. They compared the two samples using 77 separate significance tests. Two of these tests were significant at the 5% level. Suppose that there is in fact no difference on any of the 77 variables between people who are and people who are not schizophrenic in the adult population. Then all 77 null hypotheses are true.
1.What is the probability that one specific test shows a difference significant at the 5% level? 5%
2.Why is it not surprising that 2 of the 77 tests were significant at the 5% level?
Well 5% of 77 trials is about 3.5. So we sort of expect making a mistake about 3 times these 2 tests could easily be just tests where we made a Type I error and really do not indicate that Schizo's are different wit regard to the measured variables.
MBS-2. Producer's/Consumer's Risk
In quality control applications of hypothesis testing, the null and alternative hypotheses are frequently specified as "The production process is working" and "The production process is not working" respectively. Because of this, alpha is sometimes referred to as the producer's risk and beta is referred to as the consumer's risk. An injection molder produces plastic golf tees. The process is designed to produce tees with mean weight of .25 ounce. To investigate whether the injection molder is operating satisfactorily, 40 tees were randomly sampled from the last hour's production. Their weights (in ounces) are listed in the attached dataset.
1.Do the data provide sufficient evidence to conclude that the process is not operating satisfactorily? Use alpha = .01.
No, p<.0001 so we will reject the hypothesis that there is no problem with the manufacturing process. In the context of this problem, explain why it makes sense to call alpha the producer's risk and beta the consumer's risk.
We might be rejecting this without good reason but chances for that would be less than .01. So there is a risk that the producer will go through the trouble of checking the process for mistakes but there is really none. That is why we call this type of error when we reject H0 but H0 is true the " producer's risk". The other error would be failing to recognize the problem although there is one, i.e. failing to reject H0 when it is in fact false the "consumers risk" in this case.,
MRB-9. Why are Large Samples Better
Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample (or the number of subjects in an experiment) on each of the following:
1.The margin of error of a 95% confidence interval
the P-value of a test, when the null hypothesis (Ho) is false and all facts about the population remain unchanged as n increases. decreases
Optional The power of a fixed level a test, when a, the alternative hypothesis, and all facts about the population remain unchanged. increases
Confidence intervals connection
WEN-7. Highway Statistics
In 1990 the average passenger vehicle was driven 10.3 thousand miles, as reported by the U.S. Federal Highway Administration in Highway Statistics. A random sample of 500 passenger vehicles had a mean of 10.1 thousand miles driven for last year. Let m denote last year's mean distance for all vehicles.
1.Perform the hypothesis test
Ho: m=10.3 thousand miles
Ha: m ≠10.3 thousand miles
at the 5% significance level. Assume last year's standard deviation of distances driven for all passenger vehicles is 6.0 thousand miles.
N=500, mean=10.3, s=6 the p-value is the area to the left of 10.1 and to the right of 10.5 which is 2*.229=.458. So there is no reason to reject H0.*Intpr
2.Find a 95% confidence interval for m.
3.Does the value of 10.3 thousand miles, hypothesized for the mean m, in the null hypothesis of question 1, lie within your confidence interval from question 2? Yes.
4. Repeat questions 1-3 if the 500 passenger vehicles sampled were driven a mean of 10.9 thousand miles last year. Reject Ho. Conf. int:(10.376,11.428), no.
5.Based on your observations in the first four questions, complete the following statements concerning the relationship between a two-tailed hypothesis test,
H0: m = m0
Ha: m ¹ m0,
at the significance level a and a (1- a)-level confidence interval of m:
i. If m ¹ lies within the (1- a)-level confidence for m, then the null hypothesis (will, will not ) be rejected.
ii. If m ¹ lies outside the (1- a)-level confidence for m, then the null hypothesis (will, will not ) be rejected.
Let us suppose that scores on the mathematics part of the Scholastic Assessment Test (SATM) in the absence of coaching vary normally with mean m = 475 and s = 100. Suppose also that coaching may change m but does not change s. An increase in the SATM score from 475 to 478 is of no importance in seeking admission to college, but this unimportant change can be statistically very significant if it occurs in a large enough sample.
To see this, calculate the p-value for the test of
Ho: m = 475
Ha: m greater than 475
in each of the following situations:
1.A coaching service coaches 100 students. Their SATM scores average y-bar = 478.
2.By the next year, the service has coached 1000 students. Their SATM scores average y-bar = 478.
3.An advertising campaign brings the number of students coached to 10,000. Their average score is still y-bar = 478.
P=.001 (s/rootn=1 in this case so we are 3 SD away form the mean)
MRA-8. SATM Confidence Interval
Give a 99% confidence interval for the mean SATM score m after coaching in each part of the previous exercise. For large samples, the confidence interval tells us, "Yes, the mean score is higher than 475 after coaching, but only by a small amount."
1.A coaching service coaches 100 students. Their SATM scores average is 478.
2.By the next year, the service has coached 1000 students. Their SATM scores average is 478.
3.An advertising campaign brings the number of students coached to 10,000. Their average score is still 478.
Hypothesis tests for proportions
MBS-7. The Pepsi Challenge
"Take the Pepsi Challenge" was a marketing campaign used recently by the Pepsi-Cola Company. Coca-Cola drinkers participated in a blind taste test where they were asked to taste unmarked cups of Pepsi and Coke and were asked to select their favorite. In one Pepsi television commercial, an announcer states that "in recent blind taste tests, more than half the Diet Coke drinkers surveyed said they preferred the taste of Diet Pepsi". Suppose 100 Diet Coke drinkers took the Pepsi Challenge and 56 preferred the taste of Diet Pepsi. Test the hypothesis that more than half of all Diet Coke drinkers will select Diet Pepsi in the blind taste test. Use alpha = .05. What are the consequences of the results from Coca-Cola's perspective?
2. phat=.56 using the tool the P(phat>.56)=.43
So we fail to reject the hypothesis that 50% of diet coke drinkers will choose diet coke. So we really cannot claim that diet coke drinkers favor Diet Pepsi.
3. the result on the other end shows as if coke drinkers could not distinguish the tastes so it looks like their preference is for the brand and not the taste.
MCS-5. Skin Improvements
Pond's Age-Defying Complex, a cream with alpha-hydroxy acid, advertises that it can reduce wrinkles and improve the skin. In a study published in Archives of Dermatology (June 1996), 33 women over age 40 used a cream with alpha-hydroxy acid for twenty-two weeks. At the end of the study period, 23 of the women exhibited skin improvement (as judged by a dermatologist).
Is this evidence that the cream will improve the skin of more than 60% of women over age 40? Test using alpha =.05.
2. n=33, phat=23/33=.7
3. Set density toll p=.6, n=33
4.Find and interpret the p-value of the test. P=.12 so at 55 level there is not enough evidence to believe that p>.6. We fail to reject H0.
The p-value tells us what is the chance that in a sample of 33 women 23 or more will see improvement if in the population 60% of the women ended up with improved skin.
MCS-7. Political candidates
To get their names on the ballot of a local election, political candidates often must obtain petitions bearing the signatures of a minimum number of registered voters. In Pinellas County, Florida, a certain political candidate obtained petitions with 18,200 signatures (St. Petersburg Times, Apr. 7, 1992). To verify that the names on the petitions were signed by actual registered voters, election officials randomly sampled 100 of the names and checked each for authenticity. Only two were invalid signatures.
Is 98 out of 100 verified signatures sufficient to believe that more than 17,000 of the total 18,200 signatures are valid?
1. H0: p=17,000/18200=.93
3. Chances to see 98% in the sample if p=.93 equal to the p-value which is p=.025 so there is only a 2.5 % chance to see this if p<=.93 that is if we have 17,000 or less valid signatures. Therefore there is a good reason to believe that we have enough evidence for more than 17.00 valid signatures based on the sample.
Repeat part (1) if only 16,000 valid signatures are required. This p-value hen
H0:p=16000/18200=.88 is much smaller so our evidence is even more significant supporting the claim that at least 16000 signatures are valid overall.
Earthquakes are not uncommon in California. An article in the Annals of the Association of American Geographers (June 1992) investigated many factors that California residents consider when purchasing earthquake insurance. The survey revealed that only 133 of 337 randomly selected residences in Los Angeles County were protected by earthquake insurance.
What are the appropriate null and alternative hypotheses to test the research hypothesis that less than 40% of the residents of Los Angeles County were protected by earthquake insurance?
Do the data provide sufficient evidence to support the research hypothesis? Use m =.10.
P(phat<.3946)=0.41 so there is a good chance to see a sample proportion as low as 39.46% or lower even if the true population proportion is.4. Not enough evidence to reject H0 at any level not even at 10% level.